text:
What should I consider when storing a light aircraft outdoors for several months?
I would like to store a Cessna outside for a period of time, close to six months. The temperature here will not exceed 30°C and will not go below 0°C.
With tie downs and a tarp, how long can I store a Cessna outside for in these conditions? Should I invest in a shed of some sort? I would prefer a specific period of time please.

Are you able to go out and start it a couple times a month? Is there a local FBO that can check on it? Maybe a pilot friend can run it around the patch every month or so? Is it flyable?

Surely there are more factors than just temperature. Things like humidity (mold), precipitation, insects (wasps), reptiles (snakes), small (mouse) and large (bear) mammals?

@RonBeyer, yes, pilot mate can start it when needed. Yes it is flyable.

I am not a real pilot, but from flight simulation I know, that after a month without operation, the oil get's black and it acts like an acid. So, be sure to remove the oil if you don't want a corroded plane.

Flying is the best thing for an airplane, rather than storing it for months as it will air out the frame and wings.

@NoahKrasser That is false, you don't want to store an aircraft (or aircraft engine) without oil in it. Oil just doesn't break down that fast, and I'm not sure I've ever heard of it "acting like an acid".

@RonBeyer I really can be wrong. And it just was an assumption to remove the oil from the engine. Maybe you know the A2A Cessna 182 for flight sim. If not: It's modelled like the real thing and in the handbook they say, that after around a month of no operation, the oil gets black and corrodes parts of the engine. I guess this is true, because the intention of the A2A C182 is to provide a training aircraft for the real world in the home flight sim. But if you say one doesn't want to store an aircraft w/o oil, then I guess it's good to start the engine every month and let it run a bit.

@NoahKrasser I own a Cessna 177B which sits unused some months (winter ones mostly), when we change the oil (three times a year at most), the oil gets sent off for analysis. It always comes back fine. My A&P just recommends we run it around the patch at least once every 45 days and keep the battery on a maintenance charger. Oil manufacturers recommend oil changes every 50 hours or 4 months. Acids in the oil (if found) are caused by piston blow-by, not degradation of the oil. In an otherwise healthy engine, you should not have fuel in the oil.

Are you in a wet or coastal environment?  You can store a "Cessna type" plane outside for a few months and be ok.  Most flight school planes live outside for years and do just fine.
Don't use a tarp.  You might damage antennas, windows, or paint.  Custom fitted covers are good but they're also expensive.
Don't just start the plane and let it idle.  That hurts the engine more than it helps.  To do any good, the engine must reach and sustain normal operating temperature until the moisture boils off.  A quick flight can accomplish this.


text:
Why is the flight path from Taipei to Tianjin not a straight line?
The flight path from Taipei, Taiwan, to Tianjin, China, does not go directly to Tianjin in a straight line.
Instead, it heads directly north, then banks left (west) before heading north again.

(Actual flight CCA188, source: FlightAware)
Possible route: PIANO L3 VIOLA R596 OKATO W13 PINOT B221 OKTUG W506 HGH W554 DOGVI A470 PIX A593 VYK VYK9A/16R (used by virtual crews on simulation networks, source: VATAware.)
Does anyone know why this is?

Possible duplicate of Why do pilots use airways instead of just "flying direct" every time?

@Simon In this case it's a bit more specific for a given route, same as with the Germanwings route.

@mins The routing via PIANO would most likely not take that turn. This is probably the DRAKE routing to SULEM. Also: vataware for routes? I wouldn't use them as a source, I know how these routes are created by some pilots :D

I posted the same question in Travel and got the following explanation:
http://travel.stackexchange.com/questions/58217/why-is-the-flight-path-from-taipei-to-tianjin-not-a-straight-line
Basically, the Chinese government operates very narrow airways that commercial flights have to use. Also, thanks for adding in the picture of the route whoever did that.

@mins I take that back. I didn't notice that VIOLA intersects with both departure routings, DRAKE and PIANO.

@SentryRaven: Thanks anyway. What would you use to find an actual route in this area?

@mins Not really many options. But the simulations folks sometimes use auto-generated routes that are not valid, so vataware is not a #1 source on my list unless I can pass that route from there through CFMU for Europe, e.g..

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This will be due to routing and the airways available. Looking at the Enroute Chart for the Taipei FIR, you will notice that there is no direct route northbound, but all routes have a north-east direction to remain clear of restricted area RCR8 (GND - FL370).

(Source: http://eaip.caa.gov.tw)
The flight most likely, as I cannot find the flightplan, departed RCTP - Taipei via the DRAKE1A or DRAKE1B SID, leading to SULEM. Thereafter it continued westbound on the respective airways within China to then join another airway north.
This is consistent with the flightprofile tracked on flightware for EVA730 from RCTP to ZBTJ.

(Source: www.flightaware.com)

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text:
Why did Airbus skip a few numbers when developing the A380?
So at the time the A350 didn't exist so I will list the aircraft not including the A350. So it goes A300, A310, A318, A319, A320, A320, A321, A330, A340 then it skips to the A380. So why did Airbus skip a few numbers?

A little punctuation goes a long way, especially after commas. :)

because airbus want largest airliner have larger number, not to be the case of b747&b757

The question of aircraft name variants have already been discussed, but I just want to add that other reasons have also been proposed:

The '8' in the 380 resembles the double decker configuration of A380.
It was named A380 because it is twice the size of A340
And of course, it is the lucky number for Chinese.

I don't know if this has been a reason for the A380, but many vehicle models are now named using "8" because China is expected to be the big market. The previous market (Occident) was more fond of the "7" found in Boeing models.

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text:
Why are ARINC labels are transmitted LSB first?
Why are ARINC labels reversed before transmitting them to destination and the LSB (less significant bit) is transmitted first:

Actual label = 0x58: 0101 1000
Transmitted as label = 0x1A: 0001 1010

Is there any particular advantage over formal transmission and label reversed transmission?
Also all systems are designed for odd parity. Why is that?

See https://en.wikipedia.org/wiki/ARINC_429#Bit_numbering.2C_Transmission_Order.2C_and_Bit_Significance

I strongly doubt there is any advantage to either. Simply historical coincidence.

Rather than opinion, I feel the answer might lie into computer science history (and thus this question might be suited for their SE)

@mins posted what looked like a great answer (then deleted it) based on the ARIN 429 Commentary. I'd encourage him (her?) to restore the answer, but just in case, I've re-linked the PDF he pointed to.

@FreeMan: I'm not happy with my answer, but I'll undelete it so that someone can use and improve. Feel free to reuse the content.

https://en.wikipedia.org/wiki/Endianness

Revised answer
The ARINC 429 frame is composed of 4 bytes (32 bits numbered by convention 1 to 32 from right to left). Bits 11 to 29, which are the payload of the frame, can represent data using different formats, e.g. binary or BCD data. The label is used to indicate the content of the data field, and which format is used, and is necessary to convert data bits into useful information.

ARINC 429 frame used in BCD format (source)
The label is usually spelled as an octal quantity (rather than decimal or hexadecimal), e.g. 010101102 will be spelled as 1268 (01 010 110) rather than 5616 (0101 0110) or 8610.
The way bits are numbered somehow makes the label the "least significant byte" and bit 1 the "least significant bit". However:

The format above is for naming and ordering information, but doesn't tell how bits are really stored in components memory.
In digital engineering, bits are usually numbered from 0 to 31, as the number actually indicates the weight of the bit: 0 is 20 (1), 4 is 24 (16). In this convention bit 0 is indeed less significant than bit 31.

The ARINC 429 data bus is composed of 2 wires and the ground reference. When an ARINC frame is transmitted on the bus, it must be serialized, which means sending bits one at a time. The transmission order is:

The label byte transmitted first, "MSB" first: 8-7-6-5-4-3-2-1.
The other bits are then transmitted "LSB" first: 9-10-11 ... 30-31-32.

So, back to your question, why not using the left to right order for the whole frame? Or at least the left to right order for the label if it's sent first?
From Wikipedia:

Like CAN Protocol Identifier Fields, ARINC 429 label fields are
  transmitted most significant bit first. [...] This notional reversal
  also reflects historical implementation details.

So let's assume there is no reason to reverse the label bits, other than historical implementations that wanted to mimic the behavior of CAN. In the French version of this Wikipedia article, it is stated that this reversal helped to read the label when looking at a frame using an oscilloscope. The MSB arriving first also appears on the left of the display (the signal on an oscilloscope generally enters on the right of the screen and scroll towards the left side as a function of the time):

ARINC 429 bit stream on a datascope (source)
I'm a bit baffled by this explanation, as engineers working with binary representations are used to read them is both directions quite easily after a short time.
Still according to Wikipedia, hardware used to serialize and de-serialize the ARINC word were actually insensitive to the order of the label bits:

ARINC 429 transceivers have been implemented with 32-bit shift
  registers. Parallel access to that shift register is often
  octet-oriented. As such, the bit order of the octet access is the bit
  order of the accessing device, which is usually LSB 0; and serial
  transmission is arranged such that the least significant bit of each
  octet is transmitted first. So, in common practice, the accessing
  device wrote or  read a "reversed label" (for example, to transmit a
  Label 2138 [or 8B16] the bit-reversed value D116 is written to the
  Label octet). Newer or "enhanced" transceivers may be configured to
  reverse the Label field bit order "in hardware."

About odd parity
It is usual in digital serial communication to use data value changes (0 to 1 or 1 to 0) to derive a clocking clue in order to ensure the receiver samples the line voltage at the right time. This is less expensive than transmitting the clock signal on a separate wire. As each bit lasts the same time, the receiver uses a local clock that is roughly in sync with the transmitting clock. To actually maintain the synchronization in the long term, the receiver uses the raising and falling edges of the signal to detect bit borders and phase-align its clock.

Resynchronization of receiver clock on front and rear edges of the signal
That works well if the delay between edges is short. This means avoiding sending long series of 0 or 1. The parity bit can help. For data containing only 0 bits (or 1 bits) the odd parity bit will be 1 (or 0), preventing long edgeless streams to occur.
In some uses of ARINC 429, the parity bit is not transmitted, and is used for data, but nevertheless it seems superfluous for sync in this case since the link protocol is 3-level with return to zero, ensuring two edges for each bit of data.

Return to zero coding (source)

As an IT guy, this answer works for me!

There's an incorrectness in this answer: You change the order you read the ARINC word (From-right-to-left to from-left-to-right), but you keep the label LSB as being the first bit transmitted. The question with ARINC-429 is that the first bit transmitted is actually the label MSB (The bit transmission order is 8-7-6-5-4-3-2-1-9-10-11... up to 32)

Sorry @mins. I don't have an answer. I just noticed that you say that label LSB is transmitted first, but, in fact, on ARINC-429 protocol, label MSB is the first transmitted bit. It's not a matter of just reading the word from left-to-right or right-to-left.

@Alex: I completely rewrote the answer. Thanks again for your comment.

I have an idea that is used in some other cases and I think that the reason in this filed is also the same. Suppose that we must take a decision from incoming code or number, which bit of number gives us more information, MSB or LSB? Of course MSB has more information especially for estimation of magnitude of the incoming number.
In some microprocessors, the high bits of instruction codes show us the instruction group. As an example, in Z80, if the first two MSB bits of instruction be 01 immediately we conclude that we have a register to register data transfer (LD r,r’) and we can prepare the hardware for executing the instruction.
Now if we need some preparation for receiver hardware before completion of incoming label bits, so we prefer that giving the MSB bits at the first.

I'm not sure that this is useful information. Much of the transmission order was driven by pre-existing standards and doesn't really take into account the type of hardware receiving the label.

Most parts of "mins" are right except minor words. And it is found the statement on the net or some tutorials of the aviation comments are not correct, which says "transmission order is 8-7-6-5-4-3-2-1-9-10-11-12-13...-32". First this expression is not precise since what the number(8,7,6...32) of the order are. Probably it is the ARINC 429 word bit position：

32|31..30| ......8 7 6 5 4 3 2 1

P.|..SSM.|...........Label

For ARINC 429 word, it is often said that MSB is left and LSB is right based on above bit position.
But this omit something. This is not applied for whole word. And what's more, MSB and LSB must be used for the specific scopes.
Usually, there is no confusion or mistakes for such omissions. But for 429 word, it is constructed with different groups.
And the MSB/LSB of label is not same with other groups.
So from my respective, don't use the "LSB of ARINC 429 word, MSB of ARINC 429 word", which brings confusion.
For the label, the MSB of the label is the right-to-left number based on above writing/reading sequence.
Specifically(based on above bit position),
MSB of the octal label part 1 is bit 1,LSB of octal label part 1 is bit 2;
MSB of the octal label part 2 is bit 3,LSB of octal label part 2 is bit 5;
MSB of the octal label part 3 is bit 6,LSB of octal label part 3 is bit 8;

32 31 30........8.......7....6|5.......4.......3|2.........1
................LSB........MSB|LSB..........MSB|LSB.......MSB
P...SSM........oct label part3|oct Label part 2|oct  Label part1

And in aviation, we often call label xxx such as Label 260(octal).
Actually label 260 is expressed in ARINC Standard Part1-17).
Obviously, like decimal numbers, octal 2 is the most significant number,
octal 0 is the least significant number(Note i don't use MSB/LSB as the single number
will be constructed with MSB and LSB). Then the label is put into the ARINC 429 word.
The octal "2" as oct label part 1 is put into the position  1 and 2.
Here it is different with other parts of ARINC 429 word.
If it is like other parts, the octal "2" should be put into position 8,7,6.
(Continue) oct "6" as oct label part 2 is put into position 3,4,5.
(Continue) oct "0" as oct label part 3 is put into position 6,7,8.
And each part of the octal number:
MSB of part 1 is the position 1, LSB of part 1 is the position 2;
MSB of part 2 is the position 3, LSB of part 2 is the position 5;
MSB of part 3 is the position 6, LSB of part 3 is the position 8(right to left).
So the octal label 260 in ARINC 429 word is like:

32|...31 30|....  8  7  6 | 5 4 3 | 2 1

.P|......SSM|......0  0  0 | 0 1 1 | 0 1

......................0....|.....6.|..2

This is the Table 6-8 of ARINC 429 part1-17.
Now back to transmission order, we often say that the label of ARINC 429 word is transmitted first
and the MSB of label is transmitted first. Such statement is right just as the standard.
In fact, based on "MSB of label is transmitted first", apparently, for example octal label 260,
see above bits arrangement, the bit number 1 of octal "2" is the transmitted first since 1
is the MSB of octal "2"(also MSB of octal 260), then 0,1,1,0,0,0,0.
IF use the bit position, the transmission order is  position 1(first transmission)->position 2, 3,4,5,6,7,8.
And for the whole ARINC 429 word transmission order is 1,2,3,7,8...32.(here 1,2 is the bit position as above display).
But why are there so many statement of the transmission order like 8,7...1 9,10....32?
I guess there may be two.
One is misunderstanding of this paragraph in ARINC 429 Standard
"The LSB and LSC of each word should be transmitted first. It may be noted that the LSB of the
word is the MSB of the label and the label is transmitted ahead of the data in each case.
This 'reversed' label characteristic is a legacy from past system in which the octal coding of
the label field was, apparently, of no significance."
The other reason:
Maybe some aviation company product does not comply to ARINC 429 standard,
but to comply to the standard, use such statement to convert it.
And back to the original question: [Actual label = 0x58: 0101 1000], in fact, the label in aviation is not spell as 0x58.
First the 0x 58, 0101 1000 is actually reading style of left-right in ARINC 429 word for the label,
i.e. reading the label like other parts of ARINC 429. This style is convenient to software programming since
maybe the driver of ARINC 429 card doesn't provide the octal style setting.
But to translate this label to ARINC standard meaning, it should be spelled per standard(from right to left), which is label octal 032.
The octal 032 is just the "Transmitted as label = 0x1A: 0001 1010". The hex 0x1A spelled as octal is octal 032.
Find the label in ARINC 429 standard, the label 032 is often used as "ADF Frequency".
Last word, the original question "why the LSB (less significant bit) is transmitted first" actually, you thought most left 0 is LSB,
but the most left 0 is the MSB of the label. ARINC standard requires that MSB of label is transmitted first for compatibility of
past system(stated in the ARINC 429 standard).

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If I am at absolute ceiling, can I climb further by trading airspeed for altitude?
If I am flying at absolute ceiling, can I still climb by trading airspeed for altitude?

How could that be, please? What is 'absolute ceiling' meant to mean?

@RobbieGoodwin: According to the wikipedia link someone added to the question, The absolute ceiling is the highest altitude at which an aircraft can sustain level flight.  The OP is asking about a non-sustained non-level-flight maneuver.

@PeterCordes Both your an Wikipedia's interpretations might be correct and still, it would behove Hitomhi to explain that in the exposition, and that in detail.

@RobbieGoodwin: Yes, agreed.  Lack of detail / explanation like that is why I didn't upvote the question.

Purely FWIW for a general reader the definitions seem perfectly obvious and inevitable and the question  seems quite clear.

Yes, you can zoom climb. While you have no excess power to sustain a climb, if you just pull back you can gain altitude briefly. You will rapidly lose airspeed and will stall if you attempt to keep climbing for long.
Some airplanes can make it higher overall by accelerating to high airspeeds at lower altitudes and then zoom climbing all the way up. This has been used in fighter jets to reach over 120,000 feet, which is far higher than the altitude the same fighter jet can reach in a sustained climb.
Near the apex of such a climb, the airplane may be essentially a projectile- the thrust is minimal if it exists at all and the airspeed can be well below the 1g level stall speed.
If you want to try this yourself without spending ages climbing, you can basically simulate it by just using less throttle. For instance, you can figure out the would-be absolute altitude if your airplane had 80% as much power as it did, then climb there at full power, dial back to 80%, find what airspeed allows you to stay level, then pull back and see what happens.
This essentially simulates the exact same thing for an identical airframe with a smaller engine. It also emphasizes that nothing special is happening aerodynamically at the absolute ceiling- it's just the altitude where the engine can't work any harder to keep you climbing.

When flying at a susstained flight at the absolute ceiling, do you have exess kinetic energy & aerodynamic forecs to pull higher?

@Apfelsaft Yes, whatever margin you have over the stall speed is excess that allows you to pull higher.

@Apfelsaft: If you were right at stall already, drag would be pretty high.  Presumably absolute max altitude is only achievable at a better lift/drag ratio, which means a higher speed and lower angle of attack.  https://en.wikipedia.org/wiki/Lift-to-drag_ratio#/media/File:Coefficients_of_Drag_and_Lift_vs_AOA.jpg .  (Drag = thrust since you're in sustained level flight, so better L/D means more lift from the same thrust.  Or the same lift in thinner air to balance your weight?)  So you'll have some kinetic energy to trade for altitude before stalling.

Hrm, or would absolute max altitude typically be at pretty high angle of attack, like close to C_Lmax (which is right at stall)?  I guess it comes down to what speed gives you the most lift from your available engine thrust (jet) or power (prop), assuming those are close to constant with airspeed for a given altitude.  And higher speeds mean more of your drag is form drag, not drag directly associated with creating lift.

@PeterCordes If you are flying at absolute max altitude, would you necessarily be flying at the AofA giving best L/D?  If you were not, you could get the lift you need to sustain flight (i.e. equal to your weight) with less drag than the thrust your engines are currently producing, which seems contradictory... If this is so, then note that the AofA for max. L/D rarely (never?) corresponds to that for C_Lmax.

@sdenham: Yes, that was my first thought, but does the shape of L/D change with IAS and/or altitude?  To get more lift at a given angle of attack, you have to go faster, and that increases drag a lot.  Ok, another graph on the same article (https://en.wikipedia.org/wiki/Lift-to-drag_ratio#/media/File:DragvsSpeed.jpg) showing speed vs. drag shows L/D max at a speed somewhat higher than stall speed, but not far  higher.  It doesn't take much more speed to let your AoA lower a lot.  That makes sense to me.  So there's maybe not a huge amount of extra energy you can trade for alt before stalling.

@PeterCordes My attempted 'proof by contradiction' that flight at abs. max. altitude must be at the AofA for max. L/D might be invalid if the thrust of the engines is a function of airspeed, in such a way that, at the abs. max. altitude, the engines are unable to produce sufficient thrust to cancel out  the drag when flying at the airspeed required to produce sufficient lift at the L/D AofA, but can do so at another speed (!) I don't know if this is a likely or even possible scenario.

@PeterCordes In reply to your reply: Lift and drag curves do vary with Reynolds number, but for the purposes of this discussion, I think the only ones that matter are those at the absolute max. altitude. Also, I agree that while these values differ, they generally don't differ by much.

@sdenham The amount of power the engine can output certainly does vary with airspeed, which is why Vy (maximum excess power) and Vms (minimum power lost) are different. Up to a point, available power increases with airspeed, so Vy is greater than Vms, so the AoA at the absolute ceiling is lower than the one that gives maximum Cl over Cd.

For some numbers- a C152 has Vx of 55 KIAS and Vy of 61 KIAS at 10000 feet. The IAS at the absolute ceiling is somewhere in the middle, perhaps 56 KIAS (Vy changes much more quickly with altitude than Vx does). The stall speed is between 36 and 40 KIAS depending on loading. So there is not a large excess of kinetic energy, but there is some.

@Chris Thank you for your reply - it answers something I was uncertain about: gaining altitude seems to require increasing lift above weight, at least momentarily, and this shows one would have a reserve of Cl in order to do so. Perhaps you could work this into your reply, as comments are ephemeral?

Vx and Vy converge at the absolute ceiling, at an airspeed that is higher than stall speed.  So yes, you can still pull back on the stick or yoke and temporarily climb a bit. Of course, you will soon be coming back down.
Also, there will be some airspeed / altitude profile that lets you do a "zoom climb" that temporarily puts you quite a bit higher than the altitude you would obtain simply by pulling back on the stick or yoke once you are in stabilized flight at the absolute ceiling.
You can read about the 91,249' altitude record set in 1958 in the F-104 Starfighter using this method at this link.  (Caution, may include ads.)  Note that engine flamed out well before the peak altitude was reached-- something to keep in mind when considering the flight dynamics involved.

At the absolute ceiling, you have no extra engine power to push you through the air any faster, so you cannot add lift to climb by pushing the power lever further forward - you are already maxed out.
Note also that at that condition, you cannot establish a climb by trimming in a bit more more angle of attack to get more lift. If you try that, you just slow down and mush along at the verge of a stall.
As pointed out below, if you haul back suddenly on the stick, you will pitch up (into a stall) and although this will furnish you with a few more feet of altitude, it's a transient effect and you will soon come back down again.
Now, just for fun, let's calculate the maximum possible height gain you can get from a "zoom climb", as follows. We'll assume a small plane mushing along at at its absolute ceiling of 12,000 feet while going 100 feet per second. Its kinetic energy is KE = $ \frac{1}{2}mv^2 $ and we will equate this to its potential energy gain, PE = $ mgh $. The $m$'s cancel and we solve for $h$, the zoom climb gain.  We get $ \frac{1}{2}v^2 = gh $ or $ h = \frac{v^2}{2g} $. For a plane going 100fps  this yields a height gain of 156 feet.
This assumes that the plane's wing is capable of converting all of its kinetic energy into height gain, which will requires enough elevator authority to pitch the plane (and the wing) up into a vertical attitude.
This also assumes that a suddenly-applied pitch-up to vertical will not stall the wing.

This "few more feet" might be an understatement. Some experimental aircraft have briefly reached space along a suborbital trajectory using that. If your airspeed is high enough at your absolute ceiling, you can trade it for a lot of altitude before falling back down.

@vsz, I was thinking more along the lines of ordinary cessnas and pipers. Should I edit?

@nielsnielsen i think you're right. the "absolute ceiling" is determiend by "best AOA" and max thrust" AND "const velocity". Hence.. you change anything, you move out of the optimum.

100 ft/sec (59 kts) seems pretty slow for a true airspeed at 12 000 feet.  (Kinetic energy scales with TAS).  For example a Cessna 172 cruises at a TAS of about 110 to 140 kts depending on model, if I'm reading wikipedia correctly.  2x as fast is 4x the energy.  At max absolute altitude in sustained level flight, would your indicated airspeed be just above stall, or would you be at a better L/D aspect ratio going significantly faster?  I'm not sure about that part, and maybe it depends on prop (constant power) vs. jet (constant thrust).

(For optimal "zoom climb" as @vsz mentions, I think you'd typically start at high speed at lower altitude, not after edging up to the point where you can't climb any higher.  But even so, unless max absolute altitude is right at the edge of stalling, with a high AoA near your C_Lmax coefficient of lift, you'd still have speed you could trade for altitude until stall + dive.)

If you reach absolute ceiling at Mach 3, there is plenty of kinetic energy left to climb further, as demonstrated by the MiG-25 derived E266.

@PeterKämpf, the thing I can't wrap my mind around is the process by which you can pull up into a vertical climb with a wing that can't take any more AoA without stalling. Is this done with engine thrust?

Supersonic wings don't stall. The E266 could still pull 5g at Mach 3. Try that with a SR-71, and it will break apart.

@PeterKämpf OMG I never knew that!

A more understandable approach may be to take out the engine for a moment and look at a Vbg glide.  Pulling back on the stick briefly reduces the glide angle, but greater drag will eventually increase glide angle, reducing glide distance.
An airplane at absolute ceiling is flying at its very best combination of available thrust and minimum drag.  Pulling the stick would only briefly increase altitude, followed by a descent below absolute ceiling.
Changing angle of attack increases lift in a generally linear fashion, but lift is lost proportionally to the square of airspeed decrease from increased drag.

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text:
Lift and Newton's laws
Newton's laws state that for every force there is an equal and opposite force. I'm thinking of an airplane like a simple system: we put x force in and get y force out. An airplane has the force from thrust and the force from gravity, I understand this. The force of gravity is constant and equals the airplane's mass times gravity. It seems like airplanes cheat newton's laws. Somehow an airplane is only generating about 1/10th the force of gravity, but it is able to use this thrust force to provide the force equal to the gravity force plus a force against the air resistance (drag). I have heard lift explained in Newtons laws as the downwash of air and how the total mass x acceleration of the air in the downwash equals the mass x gravity of the airplane. In short, the airmass acceleration generated by the propeller < than airmass accelerated by the wing: how does this make sense? What am I missing?

See the duplicate question. Another way of thinking about this: a car doesn't have the power to climb a near vertical wall, but it can (slowly) ascend a gentle slope.

I read the other one. I'm asking the question in a different way

It would help to explain how your question is different more than just saying that the answers on the other question aren't helpful to you.

You might want to read this as an introduction to the physics.  There is no requirement in physics for thrust, derived from the force applied by propulsion, to equal or exceed weight.  Consider one of the plans being considered to divert asteroids away from a collision with the Earth by attaching solar sails to them.  The sails produce very little thrust in comparison to the weight of an asteroid but over time can accelerate them enough to alter their orbit.  Newtonian physics are alive and well in all aircraft.

@DanielCaoili You're asking essentially the same question, though. As my answer on the other one says, the important part to notice is that thrust is countering drag, not gravity. The air hitting the aircraft is redirected downwards, which causes and equal and opposite force on the aircraft. The force is point mostly upwards and a little bit backwards (relative to the airplane.) The thrust only needs to counter the backwards part. The upwards part is equal to the weight of the aircraft (when the aircraft isn't accelerating vertically, that is.)

@foot yeah when I read the apparent duplicate question's top answer, in my mind, I felt like all he was saying was "well lift causes the airplane to go up, drag pushes it back and pressure causes lift".


text:
Do Rolls-Royce Trent 700s (on the Airbus A330) automatically spool up when selecting *only* idle reverse?
I recently rode in the jumpseat of an A330 during a flight. Quickly after touchdown, the PF selected idle reverse, and no further. As soon as he did so, I could hear the engines begin to spool up, and saw the EPR indications rise, before the engines went back to idle reverse about 3 seconds later. My question is, do the engines automatically begin to spool up as soon as idle reverse is selected in order to more rapidly provide maximum reverse thrust if that setting is subsequently selected?
If so, this is unusual, as all other engines will remain in idle until further reverse thrust is selected. I looked in the FCOM and FCTM, and asked the pilot flying himself, but could get no answer. The engines were Rolls-Royce Trent 700s.

When you pull the reverse levers up there is a physical detent or stop for idle reverse. It's easy to overshoot the idle detent or stop, so the engines get a short small reverse thrust input that then goes back to idle, as you pull them up, overshoot a bit, then ease them down into the detent or stop.

@JohnK On the A320 the reverse levers will actually only get you into idle reverse, you then have to pull the main thrust levers backwards for more reverse thrust. But on the A330 Airbus went back to the more traditional design (with a detent for reverse idle) for some reason.

Sounds like the 320 was an attempt to replicate turboprop power lever operation maybe, for pilots moving up from TPs. On the CRJ200, the reverse idle detent was very difficult to find on the way up, and you'd almost always overshoot a bit. Fortunately, the screwjack driven cascade reversers took about a week to deploy, so it didn't matter so much.

To anyone who is interested, I found the answer. When idle reverse is selected, it will enter "elevated idle" mode for a few seconds to more quickly provide full reverse if that setting is selected. If the reverse thrust levers stay in the idle reverse detent, however, they will change to "normal idle" mode from "elevated idle" mode. This applies to the A330-200 with RR engines, don't know if it applies to other variants. Found it in the FCOM.

To anyone who is interested, I found the answer. When idle reverse is selected, it will enter "elevated idle" mode for a few seconds to more quickly provide full reverse if that setting is selected. If the reverse thrust levers stay in the idle reverse detent, however, they will change to "normal idle" mode from "elevated idle" mode. This applies to the A330-200 with RR engines, don't know if it applies to other variants.
Found it in the FCOM:

Reverse idle
Minimum reverse idle:
The minimum reverse idle setting is the nominal idle mode when thrust
reverser lever is at the reverse idle detent (de-activated when
elevated reverse idle is active). The FADEC selects the minimum
reverse idle setting when:

The aircraft is on ground
The flight crew selects the reverse thrust, via the thrust reverser lever.

Elevated reverse idle:
The elevated reverse idle setting enables to rapidly accelerate the
engine from reverse idle to maximum reverse thrust. This mode is only
available for the first seconds of thrust reverser deployment and is
higher than minimum reverse Idle.


text:
Could a Cessna Caravan be outfitted with Wi-Fi?
From what I understand, both ATG and satellite Wi-Fi require substantially large antennas to send and receive signals in large commercial airplanes.
Could a system like this be outfitted on a Cessna Caravan, and how does the fact that the airplane flies lower and slower than larger jets influence that?

Ask Starlink?  The terrrstrial antennas are not that big...

Most Caravans won't venture above 10000ft or far from land most of the time, so just use your mobile phone.

WiFi is nothing more than a set of protocols for transmitting Ethernet data packets without wires. Your cell phone can create a WiFi hotspot, so if your cell phone has power, then creating one in a Cessna Caravan or any other aircraft is trivial. If you have 120v power available, any WiFi enabled router can be plugged in, and the network configured on it will start working. Are you looking for broader Internet access? That would take some sort of external signal processing, but the same cell phone shouldn't have a problem talking to terrestrial towers at low altitudes & slow speeds.

@CatchAsCatchCan There are more limitations to this that there may seem.  You'll typically get good LTE or 5G performance very close to the ground.  But above a few hundred feet AGL the problems compound.  Your phone will be "seen" by many towers simultaneously and you will seem essentially equidistant to quite a few of them.  Both your phone and the system will often recognize this and say "not so fast, sea biscuit."  This is why you'll often see multiple bars of data service and a 4G or 5G indicator, but still have no data connection.

@MartinJames The recommended Starlink antenna for mobile applications is a 20" x 20" flat panel, that has to be external to the fuselage, so it becomes a significant approvals problem.

Bendix King offers a TSO'd Inmarsat internet transceiver for the Caravan called an AeroWave 100, so the shortest answer is "yes."  It's not inexpensive, you're looking at an installation north of 30 to 40,000 USD before WiFi.  And you are correct, the antenna is the hardest part to deal with.

What about Iridium GO!? https://www.iridium.com/products/iridium-go/

@Someone Could be doable, but in most metal aircraft you'll really want that antenna outside the fuselage and are like staring down TSO/STC approval.

@Someone Iridium GO! provides short burst data and voice calls. It essentially does text messages, but no true internet access, like OP is likely imagining. They make a suction cup windshield antenna designed for aviation applications.

@user71659 oh, okay. I thought it gave Internet access.


text:
What is the purpose of quickly moving the speedbrake lever on the 757 during rollout?
I found two videos where the PF quickly moves the speedbrake lever during rollout on the 757. Note that they are already deployed. Here are the links: video 1 and video 2 (out of view). What is the purpose of doing so?

On B757, B767, B777, retracting the speed brakes during the landing roll is a technique some pilots use to disarm the Auto Brakes. I haven’t flown the other Boeings, but they probably also work this way.
It is especially useful if your Auto-braking is coming on a little too strong and you want to extend your rollout to the next exit.
It is easier than reaching all the way over to the right side of the panel to turn them off.
Pushing hard on the brake pedals will also disarm the Autobrakes, but that is the last thing you want to do when trying to extend your rollout.

Autobrakes 3 or 4 gives a pretty good deceleration rate, and while applying the brakes will smoothly disarm them, it will add deceleration and further the delay the rollout to a high speed exit. Moving the speedbrake handle just a little bit also smoothly disarms the Autobrakes.

And that's why generic statements seldom apply to specific circumstances.

Perhaps "trick" is the wrong word? For example when driving a car, there are a number of ways to disable the cruise control.  All competent drivers know these different ways, and, they use them appropriately.

Agreed. I just edited “little trick” to “technique”.


text:
Will ice on an airfoil work as a vortex generator?
Vortex generators are little fans that generate vortex to attach flow to an airfoil and delay stall. If frost or ice are formed on an airfoil, technically they also generate turbulence and vortex,I am not sure if frost and ice also works like vortex generators on the wing.

Little fans?  Every VG I've ever seen is fixed to the wing, with no moving parts.

IF the ice were to form in the shape of a VG, then it might stand a chance of being useful, but that's highly unlikely.

No. A vortex generator protrudes up into the free stream, and the horizontal tornado it creates circulates partly in the free stream and partly in the turbulent boundary layer. Spiralling free stream air is mixed into the turbulent boundary layer, adding more energy into it and helping to delay separation
A VG will raise stalling AOA by several degrees, say 4-6.  A leading edge slat achieves about double that, up to about 10 degrees.  VGs are sometimes called "the poor man's slat" when installed to improve takeoff performance and low speed behaviour.
Frost does the opposite. It just disrupts the flow by tripping the laminar portion into turbulent at a much lower angle of attack.  Frost or light rime ice on the leading edges will lower the stalling AOA from, say, 15 degrees, to 9 or 10 degrees.
It's not just high performance wings. People have crashed Piper Cubs and Cessna 150s with a little bit of morning frost on the wings, that the pilot couldn't be bothered to clean off.
There have been aircraft with leading edges with a very rough texture designed to work like turbulators, that seem to be successful at it, but in the case of frost or ice with any kind of roughness, expect it to be bad news if you allow AOA to get very high at all on any normal airplane with a smooth, non-slatted leading edge.
You would be out of your mind to take off in a jet with no leading edge devices and frost on the wings (slats when extended make the wing pretty much immune to light frost); I know of several Challenger and CRJ crashes from doing this.
If, say, a hijacker put a gun to your head and forced you to take off with frost on the wings, and you had no anti-ice to melt it off, you would want to accelerate well above normal rotation speed, rotate very slowly, and let it accelerate right away to climb out at a higher than normal climb speed. Whatever it takes to keep angle of attack as low as possible.

achieving some of the effect of having the boundary layer laminar. - really? Isn't it more like a super-turbulent boundary layer?

A spiraling stream that is moving is preferable to a completely chaotic layer that is semi-stagnant no?  I understood there was a drag reduction from this effect that compensates for the energy consumed to start the air spinning, which is why VGs have a more or less neutral effect on cruise speeds.

You can hardly call a turbulent BL stagnant. The important part is the speed gradient at the wall, which increases from laminar to turbulent and then again with VGs. In the end, with VGs you take energy from the free stream and bring it close to the wall. Hardly something you do to avoid drag.

Wall? Skin surface?

What's a better way to end that paragraph?

Yes, wall = skin surface. Boundary layer scientist lingo.

last sentence of first paragraph proposal: Spiralling free stream air is mixed into the turbulent boundary layer, adding more energy into it and helping to delay separation. There is too much crossflow in the vortex to make that flow even remotely resemble a laminar BL. It is really producing a turbocharged turbulent BL.

Done thanks very much

The first question is less "will it?" and more "can it?" delay stall, where the answer to the latter question is yes! There is research abound studying roughness effects like this. Generally the results depend on roughness size and location but results like the following are common: 1) The zero-lift drag coefficient will increase due to increased skin friction, and 2) at high angle-of-attack lift increases and drag decreases due to delayed separation.
The transition from laminar to turbulence is not uniquely accomplished by the "normal" vortex generator geometry as the other answer suggests, and all kinds of research has been done to investigate the effects of various other roughness and element geometries that accomplish this.
The answer to the question of "will it?" though of course is: it depends. But if were to analyze the characteristics and distribution of frost on an aerodynamic surface, you should be able to apply existing results such as [1] to determine what effect it will have. And in some cases it may result in desirable changes, and other cases not.
In the real world, however, no pilot is going to look at an icy, frosty wing, characterize it, and make an accurate determination as to how the aircraft aerodynamics and flight characteristics will change, and whether theses changed characteristics were advantageous for the flight operation you were conducting.
And even if you did determine the changes were desirable, you're still required to comply with (in the case if Part 91 operations, although I'll let you dig up the reference for other parts) 14 CFR 91.9 which states:

Except as provided in paragraph (d) of this section, no person may operate a civil aircraft without complying with the operating limitations specified in the approved Airplane or Rotorcraft Flight Manual, markings, and placards, or as otherwise prescribed by the certificating authority of the country of registry.

And I'd expect in your aircrafts POH/AFM you will find language stating that all frost, ice, and snow be removed from the aircraft. Hence, you've got to remove even advantageous frost from your aircraft to be legal.

Chakroun W, Al-Mesri I, Al-Fahad S. Effect of Surface Roughness on the Aerodynamic Characteristics of a Symmetrical Airfoil. Wind Engineering. 2004;28(5):547-564.
https://journals.sagepub.com/doi/10.1260/0309524043028136


text:
Why do piston engines have a lower fuel consumption with higher temperatures?
It is very common to find in the AFM/POH, Section 5 of most piston engine airplanes (C152, C172, PA28, BE76, etc) charts that indicate the fuel consumption for cruise/economy power settings.
You will notice that in all cases, for a given set of pressure altitude, RPM and manifold pressure (if constant speed propeller) the fuel consumption will depend on the OAT. The question is: why does the fuel flow (in PPH or GPH) decrease for higher temperatures?
I should make sense to me that higher temperatures (hot atmosphere) would decrease the overall performance of the engine, as we all know, since the air is less dense which directly impacts in the engine's operation. If the engine's performance is degraded in a hot day, why does it burn less fuel?
All manufacturers publish similar results, but I can't figure out a physical explanation to it. If you could point out an official reference in any textbook I would really appreciate.

You still need to maintain the Fuel/Air mixture to get the optimal burn. If you are running hot and there is less air (oxygen) for a given cycle, it follows that you would need less fuel to maintain the same a/f ratio. It may be less fuel but it is also producing less power too.

Please define "better". I guess all you see is a lower consumption at the same power percentage, but the reference for that percentage goes down with rising temperature, so the consumption/power ratio gets actually worse.

I just modified the title to properly address the question: "better fuel consumption" I meant "lower fuel consumption".

I don't have a reference, but the physics behind is not that complicated.
The key is that you have constant RPM, constant pressure altitude and constant manifold pressure.
A motor sucks in lots of air, but since the throttle restricts the air flow in the intake, the air pressure between throttle and motor is low. This is the manifold pressure1. At idle, i.e. almost closed throttle, the pressure is very low, while at full throttle, it ideally equals ambient pressure. And in turbocharged motors, it is even higher than ambient pressure.
Finally, manifold pressure together with RPM and displacement of the motor is a measure of how much air volume is aspirated by the motor per time interval.
Since you keep everything constant, this volume also is constant. But warmer air is less dense and contains less oxygen molecules per volume.
Less oxygen molecules means less fuel can be burned. This is why fuel consumption is lower.
But it also means less power. To compensate, the throttle could be opened a little more to let in a little more air volume. This would also increase the manifold pressure.
But if the throttle is already fully opened to get max. power, you can't open it further. So, max. power is lower at higher temperature.

1) NOTE: For simplicity, I'm speaking about absolute pressure here. The gauge in the cockpit displays the under-pressure with respect to ambient pressure, so a high displayed value means a low absolute pressure. And since ambient pressure changes with altitude, manifold pressure does, too, even if the displayed value is constant. The benefit of this is that the displayed value is a good measure of the power with respect to the max. power achievable at that altitude.

maybe you should add that a higher ambient temperature means that the thermodynamic cycle is less efficient. The physics is quite the same as that for turbo machines. So relative to power, fuel consumption even goes up with temperature.

The problem is, those performance tables show that the airspeed is the same while the fuel consumption and the power is lower. This means that the same amount of work is performed in the same time interval, thus the same net power is output by the engine with better energy efficiency (at least at first glance, fuel consumption), while the thermodynamics tells us otherwise.

@RedGlyph, I don't see the question mention same speed. Speed will be lower, but in GA you usually have defined cruise power and the speed is whatever it is.

@JanHudec It's an old question so I'm not going to read all that again, I think it was a side comment to the more general consumption issue. I can just say that any manual I've seen provides those tables in function of the airspeed, which makes sense because the important point is fuel consumption when you prepare a flight.

It's much more fundamental then that, and also much simpler.
These fossil fuel engines work by burning a fuel air mixture in a confined space. A fixed amount of burned fuel releases a fixed amount of energy. If you extract energy by cooling, then there is less energy left over to do actual work. (If you cool the hot gasses, the pressure gets lower).

Welcome to aviation.se!

You start well, but the second half you wrote makes no sense: the power stroke happens so fast that there is no appreciable cooling. And in fact, the engine is more efficient when the outside air is colder. The problem with warm air and confined space is that warm air is less dense, so less of it fits in that space.

Let's try it from the aerodynamics side.  Hotter air at a given altitude is thinner. It's the same thing as flying higher:  more TAS for the same IAS.
As long as you do not overheat, and your engine/prop can develop enough thrust, the same benefits are gained as with flying higher: better miles per gallon.
Note the airspeed on the Cessna 152 fuel consumption chart is listed as KTAS.  So as you move across the chart from cold to warm, similar KTAS is lower IAS.  It is the same as you read fuel consumption (Gallons per Hour) at higher altitudes for the same TAS.
Notice that poor prop can't even make 2400 RPM at 2000 feet and 20 degrees colder than Standard Temperature.  Its the same thing for your wing and fuselage: trying to push through air at a higher density.
From this we can gather that flying fish are on the right track!


text:
Can I execute an ILS approach using a NAV CDI with LOC signal only?
I have been flying a small Cessna airplane that has a COM/NAV NARCO MK12 equipment with CDI (lateral needle only).
This CDI does not incorporate a horizontal needle to provide vertical guidance, so there is no glide-slope indication for ILS.
Last time I tuned the LOC frequency of an ILS procedure and the CDI captured the LOC signal. The audio panel also captured the O M I markers.
My question is if it is legal to execute an ILS approach "with LOC" only limitations, as published in the approach plate (since the airplane does not have any GS information).

If you're asking what's "legal" then please tell us which country or regulations you're asking about. It would also be helpful to know a specific approach that you're interested in, so that we can see exactly the scenario that you explained.

No, you cannot execute an ILS approach without an ILS glide slope indication as the primary form of vertical guidance.
Even if you could supplement it with alternatives (WAAS, GPS software, etc), they are not recognized as the appropriate standard of vertical guidance required for an ILS approach.
Many airports that have an ILS also have LOC (localizer-only) approaches.  If such an approach is published, you could certainly fly it with only the use of the localizer, but then again, it is not an "ILS approach" it is a localizer approach.
The localizer -only approach will have different minimums which will be based on your altitude at a published minimum descent altitude rather than a decision height.
If no localizer approach is published, then you'd need to find another method or airport.

Often the localizer-only approach is published on the same plate as the ILS. In the US these are titled something like "ILS or LOC RWY 30".

Yes, thank you, that is a good point to include.

Frequently localizer only approaches use another ground-based source like a VOR cross-radial to indicate step-down fixes, so it could be challenging to fly with only one Com radio.


text:
Is Qantas holder of the world’s longest commercial non-stop flight?
Is it Singapore Airlines or Qantas that is currently holding the record for the world longest non-stop commercial flight?

I flagged this as off topic because it would be better on travel.SE

I assume you are referring to the recent Qantas flight from New York to Sydney:

Australian carrier Qantas has completed a test of the longest non-stop commercial passenger flight as part of research on how the journey could affect pilots, crew and passengers.
The Boeing 787-9 with 49 people on board took 19 hours and 16 minutes to fly from New York to Sydney, a 16,200-km (10,066-mile) route.

(BBC)
This flight was primarily a test flight to study impacts of ultra-long haul flights on passengers:

Qantas has announced three ultra long-haul research flights to gather new data about inflight passenger and crew health and wellbeing.

(Qantas News Room)
Because it was just a test flight, this does not count as regular service, which would not make sense with the current aircraft anyway:

No commercial aircraft yet has the range to fly such an ultra-long haul route with a full passenger and cargo load, Reuters news agency reports.
To give the plane sufficient fuel range to avoid re-fuelling, the Qantas flight took off with maximum fuel, restricted baggage load and no cargo.

(BBC)

The current record for longest regularly scheduled flight is indeed Singapore Airlines flight SQ21 between Newark and Singapore according to the Wikipedia List of longest non-stop flights (retrieved 2 Nov 2019, based on great circle distance, thanks to Michael Seifert for pointing this out).

The record for the longest ever flight with a commercial airliner however is still held by a Boeing 777-200LR:

In a trip that included two sunrises, a new long-range Boeing 777 flew more than halfway around the world nonstop on Thursday, breaking an aviation distance record set by a B-52 bomber 43 years ago.

New York Times
This flight was primarily a publicity stunt resulting in a Guinness World Record. It was never meant to be used commercially since they intentionally flew a longer route.

For future readers, note that Wikipedia has a "Longest flights" page that is generally kept current.

@MichaelSeifert Good point, I removed the BBC image and replaced it with a link to that table. Thanks for the hint :)

It is difficult to determine the longest actual flight route as, depending on the route chosen by the pilots, the distance varies. The route from Qantas, JFK-SYD measures 16013 kilometers. This is roughly 600 kilometers longer than the route SIN-EWR operated by Singapore Airlines.
Looking at the data provided by FlightAware for this flight operated by Singapore Airlines, it can be seen that the flight from October 31st, has flown 16850 kilometers. Thus a longer route than the scheduled route by Qantas.
In general the distances for the longest route are determined using the great circle distance measurement. Then the route flown by Qantas is the longest route, which in practice will be actual longer than the 16013 kilometers


text:
Why do I missing 1090 Rebroadcast Traffic (ADS-R) Targets from Ground Station as part of 978 Uplink?
For context I have an ADS-B 1090 OUT and 978 IN. I also have a Stratus portable Dual channel receiver in my plane, so I have the option of comparing the traffic between systems.
What I’ve been noticing is that lots of 1090 targets that I can see on Stratus connected iPad, I can never ever see it on my display connected to 978 IN avionics. My understanding is that my setup should:
A) Receive direct reports from aircraft equipped with UAT/978 OUT hardware. (Air-to-Air) (Working Good)
B) Receive ADS-R rebroadcast traffic from ground stations, for 1090 targets. (Not receiving)
C) Receive TIS-B traffic from ground stations, for Mode-C targets that aren’t participating in ADS-B network. (Good)
So my question is, why do scenario B seem not to be working for me?
I pulled ADS-B Performance Report for my flights and I see:

OUT capability is correctly listed as 1090 and IN capability is listed as UAT. This should indicate the Ground Stations that I need 1090 targets as part of rebroadcast (ADS-R) to me. I contacted FAA on it and they agree too

Any thoughts? is it something I am missing in the ADS-B OUT configurations or could it be Ground Station issue?

Have you checked your configuration against the info here: https://www.faa.gov/nextgen/equipadsb/capabilities/ins_outs/ ?

@Gerry - Yes I did. The ADS-B OUT configurations are correctly listed in the PAPR report as well. OUT - 1090, IN - 978

Ground stations only send ADS-R/TIS-B messages for targets within a certain distance horizontally and vertically of ADS-B targets with the CDTI (Cockpit Display of Traffic Information, aka "ADS-B In") flag set. Assuming your CDTI flag is set correctly, the targets you're missing probably aren't close enough to qualify.


text:
When rearranging the lift force equation to solve for lift coefficient instead of lift force, why is the numerator often given as F not 2F?
the standard description of the Lift force affecting the lift surface is:
$$
F={\dfrac{1}{2}}\rho v^{2} SC_{L}
$$
the basic rearrangement gives the following definition of the lift coefficient:
$$
C_{L}={\dfrac{2F}{\rho v^{2}S}}
$$
why is the above expression met so rarely? with the standard skipping the 2 multiplier?
$$
C_{L}={\dfrac{F}{\rho v^{2}S}}
$$
I've checked 30 scientific papers, those measuring the lift\drag coefficients.

28 papers have had F
2 papers have had 2F

Could it be related to some assumptions about S, say S in one case is for one wing and in the other case, for the total wingspan ? It really depends on the context. But to me there should definitely be a 2 factor

Are you sure that there’s no 0.5 in the denominator ? Can you give a reference to confirm?

The definition of the coefficients differs between countries/regions. The factor ½ is used in most of Europe and in Russia, because it is part of the dynamic pressure equation (in metric units). In the USA working with USA units, the 2 is incorporated in the coefficient.
As for instance mentioned in Principles Of Helicopter Aerodynamics, section 2.5 by J. Gordon Leishman, Cambridge University Press:

It is important to note that the US customary definition of the thrust, torque and power coefficients is different to that used in some parts of the world (mainly in Britain, most of Europe, and Russia), where a factor of one half is used in the denominator..

Not contradicting anybody, just note the(1/2) factor in the lift force comes directly from  Bernoulli equation

Really? Anderson's Fundamentals of Aerodynamics, which I consider quite American, uses the "normal" (to me as a European) definition of q = 1/2 rho V^2 and C_L = L / (q S), which implies C_L = 2 L / (rho V^2 S).

@Raketenolli Does Anderson use metric units?

@Koyovis both, but he stresses the fact that he uses a consistent set of units. I just saw your edit, how is 1/2 part of the dynamic pressure equation in metric units - but (I guess that's what you mean) not in US customary units?

Doesn't matter whether units are metric or US.  Apparently, OP's noted discrepancy is due to a typo.  The lift coefficient is dimensionless, of course.  And 1/2 in the denominator translates to 2 in the numerator.  Evidently, somebody forgot the 2 in setting up the press galley and missed the error in proof.

The total aerodynamic force acting on a body is usually thought of as having two components, lift and drag.
By definition, the component of force parallel to the oncoming flow is called drag
and the component perpendicular to the oncoming flow is called lift
It is convenient to express the induced drag as an expression having similarities with that of the lift, thus by taking the ratio of lift to drag the « 2 multiplier » disappears, but considering solely the lift coefficient the expression you are giving is correct.


text:
How much thrust should be applied when you set the autopilot to a specific altitude?
After a plane takes off the pilot usually sets the autopilot to the altitude of the flight level they are about to cruise at, e.g. 30,000 ft, right? When that is the case, how much thrust should be applied during further ascent and then at the cruise altitude? Or does the autopilot do this itself as well? How would the autopilot behave if you ascend at 100% of thrust or at a thrust too low?

The autopilot commands the pitch and roll and uses elevator and pitch trim as well as ailerons and rollspoilers for this.
An optional auto-throttle system changes the thrust or power of the engine.
Some aircraft, not all, have an auto-throttle system, for example: Boeing 737, 747, 757, 767, 777, 787, Airbus A320 family, A330, A340, A350, A380 all have this. Other aircraft such as the Dash-8 Q400, Cessna 172 and most turboprops and piston and general aviation aircraft do not have an auto-throttle system.
In aircraft with the auto-throttle, when the system is turned on, the A/THR changes the thrust between idle and climb thrust (not full 100% thrust). So during the climb the auto-throttle advances thrust to climb thrust and keeps it there until the new altitude target is captured. During the climb the autopilot pitches up and down to maintain the set target speed. Then, upon reaching the target altitude the auto-throttle switches back to maintaining the airspeed and the autopilot pitches up and down to maintain the altitude. During the descent the auto-throttle system reduces thrust to idle.
In aircraft without an auto-throttle system, the autopilot can only change the pitch of the aircraft and you as a pilot need to feed in power for climb or reduce power for the descent. Otherwise the aircraft wont climb or descent. These autopilots may still have a flight level change mode where the autopilot maintains a set airspeed target and pitches up and down as you increase or decrease power but you still need to control the engine manually. And once the target altitude has been reached you also need to change thrust or power to fly at the desired speed.

Does the auto-throttle have to be turned on seperately or does it turn on when you turn on the cruise altitude? If the former, what do you turn on first, the altitude or the auto-throttle? And at what altitude do you usually start ascending / end descending on autopilot?

@Giovanni All of those questions depend heavily on the airplane type, the policies of the company, and the preferences of the pilot. The autothrottle is usually a separate control from the autopilot, though exceptions exist. You turn them on in whatever order the checklist for that specific plane says. For big commercial planes, the autopilot and autothrottle are typically turned on a few seconds after takeoff, and remain that way until a few seconds before landing, but they can always be turned off if needed. For smaller planes, the pilot decides.

There was an autothrottle STC add-on for the CRJ200 corporate version that I've used a few times. I didn't like it because the thrust lever friction controls were removed and the AT servo clutches remained engaged full time, the clutch slip force being your one-size-only friction level, which was too high for my liking, especially on approaches.

@HiddenWindshield I mean the usual passenger jet airliners. You put in the altitude and set the switch so that the autopilot goes there. Do you immediately afterwards have to turn on autothrottle or otherwise what throttle is best so that the plane stays at the autopilot altitude?

@Giovanni There's no such thing as "usual passenger jet airliners" in this case. A 737 is going to have a different procedure than an A320. Delta Airlines has different procedures from Emirates Air. Also, what do you mean by "have to"? If you're taking a technical perspective, you don't "have to" use the autothrottle at all, any more than you "have to" use the cruise control of your car. If you're taking a procedural perspective, you "have to" turn it on when the manufacturer/company policy says you "have to".

@Giovanni, you seem to have a misconception that throttle is used to hold altitude, it isn’t. The auto pilot modulates pitch to hold altitude, power is used to control speed.

@HiddenWindshield I doubt that pilots would want to fly manually all the time at a certain flight level. So yes, they have to use the autopilot.

@Giovanni In my comment, I was talking about the autothrottle, not the autopilot. And, no, you don't have to use either one (unless, of course, required by procedure). You can drive for hours without an autodriver, you can fly for hours without an autopilot.

@HiddenWindshield Airline pilots need to fly at the very flight level ATC has told them. E.g. exactly at 30,000 ft which is hard to control manually. And concerning autothrottle, that was a question of mine. Obviously they don't need to use it for the autopilot to hold the altitude.

@Giovanni I am a pilot, and I can tell you, it is not that hard to manually hold altitude. I've flown a plane with a broken autopilot for over an hour, and never once busted an altitude restriction. Autopilots are only required in areas of reduced vertical separation minimum.

@Giovanni Usually the auto-throttle is armed before takeoff and is engaged after takeoff at around 1000ft or so (depends on thrust reduction height, airplane and operating procedure). The autopilot is typically used from 400ft above the ground until just prior to landing. Once auto-throttle or autopilot are engaged they remain on. You can still select new modes like climb, descent or altitude hold and the auto-throttle and autopilot handle all that for you. At any time during the flight the pilot can take manual control.


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What would it take for a balloon to reach the thermosphere?
So far the altitude record for a balloon is BU-60-1's 173,900 ft (53 km). What would it take to reach the thermosphere, i.e. more than ~280,000 ft? How likely is it that eventually one day a (hydrogen or helium) balloon will reach the thermosphere? What effect would the different composition of the atmosphere there have on the balloon and its ascent? How big would a balloon need to be to reach the thermosphere and what material would be best?

Strangely I have some serious domain experience here, I have in the past been involved in commercial/research balloon operations to about 130,000’ at "industrial" scale, involving tens of thousands of commercial and military launches.
The answer to "how likely" is "almost certainly not."
Newtonian reaction, or mass ejection, (what we'd call a rocket) is really hard to do down low because at launch you are lofting the payload weight plus the much larger weight of all the fuel needed for the rest of the flight.  But you get lighter as you go.  The engineering problem gets easier as you climb.
Balloons instead produce "lift" by means of buoyancy.  My vehicle will weigh just as much at the surface as it weighs at 150,000 feet.  Neglecting envelope and gas weight, if my payload weighs 6 pounds, I have to produce >6 pounds of buoyant force at the surface, AND at 40,000 feet, AND at 150,000 feet, AND at 200,000 feet.  Opposite the rocket, the engineering problem compounds geometrically with increasing altitude.
My balloon has to displace whatever volume of air is required to create a 6 pound delta.  The size of 6 pounds of air at the surface is nothing at all like the size of 6 pounds of air at 100,000 feet, which is again nothing at all like the size of 6 pounds of air at 200,000 feet.  So the balloon only climbs when it expands in rough proportion to the air density lapse rate.  My 4-5 foot diameter balloon on the surface expands to about 40 feet at 100,000 feet.  It will be profoundly larger at 200,000 feet.  As air density approaches zero, the size of the balloon envelope approaches infinity.
A pound (or any mass) of air at sea level is 68 times larger in volume at 100,000' and...  wait for it...  4,200 times larger at 200,000'. That's a lot of expansion.
And unlike a rocket, I have no orbital velocity to create a centripetal force that offsets gravity.  My balloon still weighs a lot at 200,000 feet.  I will eventually need the envelop to expand to the size of a domed stadium, and then a small city, and eventually, the planet, to keep climbing.
So there's the limiting factors:

I need to have meaningful air density to displace in order to create a buoyant force.  By definition, the thermosphere is where I have essentially run out of air density.  At the thermosphere, air becomes non-molecular and you have individual unbound atoms of oxygen, nitrogen, and helium, and you can't produce meaningful buoyancy.

I need a material that expands to potentially hundreds and then thousands of feet of diameter.  And every pound of balloon material I have to add to make a more elastic balloon envelope means more weight I have to loft.  The balloon will only climb so long as it can expand.  I have three choices:  (a) If the material structurally constrains expansion to prevent bursting, it will hover at its terminal expansion altitude, and then will plummet seconds after the balloon goes into the earth's shadow at night and starts cooling from the lack of solar radiation.  (b) It can expand infinitely.  Or (c) it eventually reaches its elasticity limit and bursts, which is what happens to all weather balloons.  Those are the only 3 choices.

I need a material that can withstand profound, hard to image environmentals.  It's going to be operating in a gas that has so few molecules that there is essentially no temperature (nearing absolute zero) while at the same time having to withstand easily 500°C of solar radiant heat, with spikes several times higher during periods of high solar activity.  So I have a monumental materials science problem to overcome as well.

Well, there's more than you ever wanted to know about balloons, but if you look at an air density table and see how big 6 pounds of air is at 200,000 feet, the impossibility of it all becomes visible.

Absolutely remarkable!


text:
Was Hawthorne C. Gray the first pilot to use a positive pressure mask?
The balloonist Hawthorne C. Gray established human altitude records twice in 1927. His ascent in May 1927 went to 42,470 ft (12.94 km) which is an altitude the FAA requires the use of a pressure demand mask today, doesn't it? Because at that altitude a diluter demand mask at ambient pressure would be unable to spare you from hypoxia even if breathing 100% oxygen.
In November 1927 Gray broke his own record by reaching an altitude of more than 43,000 ft where you definitely have to be in higher-than-ambient pressure, don't you? He died in this flight, but his death occured when the balloon crashed because his oxygen supply ran out during descent though still too high, rendering him unconcious.
I'm unsure what kind of oxygen supply Gray used during his flights, but it seems to me he was the first human to use oxygen under positive pressure. All subsequent higher flights (such as those of the Piccards) were either in fully pressurized gondolas/cabins or in pressure suits.

Without knowing the method of his oxygen supply I don't think the claim can be made.
He may well have been hypoxic depending on the lenth of time with whatever suplemental Oxygen he was using at altitude.
On the second attempt he almost certainly was hypoxic although technically 'alive' when he crashed. The determination that he ran out of oxygen "on the way down" is spurious. Maybe he was hypoxic at the top and he didn't start down in a sufficient time frame for his supply because of that hypoxia.
One thing we know is that every person reacts differently to the state of hypoxia. It is also a sliding scale not an either or state.
It could be likened to getting drunk without drinking provided there were some means to teleport alcohol directly into someones stomach. They would be getting more drunk but with no external reference they would not be able to determine how much imparement they had suffered. The sad thing is, alcohol uptake does that itself for drunk drivers.
On the alive statement,
Alive doesn't mean coherent or even conscious. It just means the heart is pumping. That can be determined purely by physiologcal means E.g. how injuries bleed.

It is confirmed that his oxygen supply ran out. It is said that his courage was greater than his oxygen supply. His last journal entry from likely just below 40,000 ft stated “Sky deep blue, sun very bright, sand all gone.” But you're right not everyone reacts the same to hypoxia and perhaps he had a long enough TUC to descend in time during his May flight. Still, this doesn't answer the question what kind of oxygen supply method he used.

My point that it is spurious to your question. The crux is knowing what supply method was used. Survival on the first attempt doesn't mean he was using a particular method that lets people survive now, only that he survived. The fact that he died during the second excursion would to me point to the fact that he was not using sealed oxygen delivery during the first session either, but if cannot be conclusivly proven either way.


text:
Why do airliners over south-west Europe cruise at flight levels with inversed numbers?
Over North America and Europe, usually when airliners go east (heading 0-179) they cruise at a flight level with an odd number (e.g. FL370) and when going west (180-359) they cruise at an even number (e.g. FL380), right?
When looking on flightradar24, all planes follow this pattern except for planes flying over south-west Europe (Iberian peninsula, Cote d'Azur, Mallorca and so, as well as Morocco): there, planes flying to the east cruise at even numbers and to the west at odd numbers. Why do planes flying over or near the Iberian peninsula have the flight levels inversed?

My guess: since the majority of traffic is North/South, they have even for northbound and odd for southbound.

@Sanchises Where they fly from/to doesn't matter, planes flying e.g. at heading 90 (east) cruise at an even number near the Iberian peninsula, when elsewhere above land the norm is to fly at an odd number east.

ATC in Florida turns the hemispherical rule 90 degrees (for all traffic) due to the dominance of north-south flows.  Maybe Spain does the same?

@StephenS Yes, maybe. I'm also suspecting Spain to have something to do with the inversion.

The above persons guessed correctly. Since a majority of flights in the area fly in north / south directions, rather than east / west, France, Spain, Italy and Switzerland generally apply the NorthEVEN and SouthODD rule. In Germany, some airways also follow this rule on their course between the Swiss and Danish airspace. This is easily seen in enroute IFR charts.
You can also check it, for instance, in the Spanish AIP, ENR 1.7.3, here:

 Spanish AIP, ENR 1.7.3  , page 3.

In the French AIP section ENR 1.3.5, you can see how they interestingly use the east / west rule for IFR flights below FL200, and the north / south rule for IFR flights at or above FL200.)

 French AIP ENR 1.3.5.

Welcome to aviation.stackexchange.com, @esther!


text:
Why does NASA's B-52 008 have a smoking engine in this photograph?
This is a photograph from the launch of an X-43 from the NASA owned and operated B-52 '008'. It appears that one engine is producing a plume of smoke in the photograph. Why is this?

Source

That's what you get when there is no cabin attendant to tell the crew this is non-smoking flight.

@RobVermeulen She was commissioned in 1959... everyone smoked back then ;)

@RobVermeulen Ladies and gentleman this is your captain speaking - I've just turned off the "No Smoking" signs. I figure if the plane's smoking why can't you? (Also can we talk about the soot marks behind those engine pods? Wash your planes, NASA!)

@Adam is right. The aircraft, NB-52B has smoke generator installed to assist in tracking.

A smoke generator has been installed on the left wing of the Boeing NB-52B Stratofortress mothership to assist in tracking the airplane.

Its use during the test of X-38 is detailed here:

Outside I could see the white trail from the smoke generator in the number-2 engine nacelle of the NB-52B approaching from the east
... I aimed my telephoto nearly directly up at the tip of the smoke trail.

As the same aircraft was used during the X-43 tests, it certainly served the same purpose.

So, paratroopers invented jumping out of a perfectly good airplane. Not to be one-upped, NASA decided to set a perfectly good plane on fire. Touche :)

My guess, and it is only a guess, is that they are using a smoke generator to make the B-52 aircraft easier to visualize.  Typically for flight tests like this they would be doing optical tracking from the ground (essentially a telescope) and this would make it easier to locate the target.  (I was going to add also for the pilot of the dropped aircraft, but the X-43 is unmanned)


text:
Could a rear center of gravity position cause an airliner to lose control when a pitch-up is commanded?
Modern aircraft cruise with the center of gravity close to the most rearward position to minimise trim drag, usually achieved with fuel in the tail surfaces. This is transferred forward before or during descent for landing.
Is it feasible that unexpected nose-up inputs could lead to partial stall with the aerodynamic center moving rearward, aggravating the nose high attitude to cause mushing? This is my suggestion that might explain the loss of AF447.
Without ASI (pitot icing) a pilot without experience of this might not recognise the problem. Further, in that attitude elevators could be ineffective and fuel transfer too slow for recovery that way.
This was a serious problem with T-tails and caused accidents to DC9, Trident, BAC1-11 and Tomahawk aircraft; maybe others.
The only solutions were stall warning and stick pusher systems, apart from streaming the braking 'chute by an RAF Victor crew.

explain the loss of AF447 - the loss is fully explained.  The only mystery is why the crew did what they did.

AF447 didn't have a loss of elevator authority. It just had bad elevator inputs (given to it by the crew.) Also, the ASIs were fully functional for the last roughly 5 minutes on AF447.

unless the speculation about AF447 is removed, to me this question should remain closed.

@Federico: Let us deal with his speculation in the answers. Closing the question just because it contains disprovable speculation is too much censorship for me!

@PeterKämpf That's a decent point. Speculation about incidents still under investigation is explicitly off-topic here, but not questions about incidents whose investigations have reached their conclusions.

@reirab I would accept it if this was a question about the final report, not a speculation about a parallel universe with alternate reality.

On virtually any swept-wing design, "wingtip stalls" are a danger. Most light aircraft are designed with a straight wing and a leading-edge contour such that the inside of the wing stalls first. This has two advantages for light craft; first, aileron control is available as long as possible into stall entry, allowing roll control and thus the ability to stay coordinated. Second, the departure of airflow over the top inside of the wing will reduce or eliminate the downwash of air over the horizontal stabilizer, which normally corrects for a slight nose-heavy CG. With the balancing force gone, the normal response of the aircraft will be to nose down, restoring a lower AOA and building forward airspeed to let the pilot recover quickly.
With a swept wing as is seen on modern jet airliners, the wingtips tend to stall first instead, because the airflow over the swept wing lowers the stagnation point and thus the effective angle of attack of rearward sections of the wing, so the wingtips at the rear will exceed the critical AOA first.
When the wingtip stalls first on a swept wing, two things happen. First, aileron control is severely affected, making it harder to stay wings-level and avoid entering a slip or even a spin. Second, the center of lift from the unstalled portion of the wing moves forward, placing the CG further behind it. An aft CG works against pilot efforts to nose the aircraft back down to recover, and in an uncoordinated stall the aft CG will keep the tail of the airplane "outwards" from the center of the spin, hindering or preventing efforts to cancel out the rotation with the rudder as the vertical stabilizer will be stalled due to the extreme "sideslip".
This is a potential problem in any airliner, where the weight is distributed along the majority of the fuselage. Any significant deviation from mean values for per-passenger or per-bag weights can place the CG too far aft. Tail trim tanks can further exacerbate this.
However, Air France 447's crash, in my amateur analysis, was not caused by trim characteristics. It was primarily caused by the pitot tube icing. This had several detrimental effects on the flight crew's situational awareness:

The disagreement of the plane's two pitot-static systems cause the flight director to unexpectedly disengage the autopilot, setting the flight crew on their heels.
This same disagreement in the pitot-static data also caused the flight director to switch from "normal law" to "alternate law 2" which among other things disabled roll and pitch/AOA limiters, as the data used to calculate those limits is no longer reliable.
Lastly, the lack of reliable pitot-static data was reflected in cockpit instrument displays. We don't know exactly what they saw because information on the status of the right side pitot system is unavailable (it was not recorded by the black boxes). However, we do know the disagreement of data caused the red-flagging of practically all information reliant on such data, including the airspeed and altimeter tapes, the flight director, velocity vector, and all secondary indicators including AOA readouts, the stall horn, etc. The flight crew would have been totally reliant on external visible references to determine the aircraft was stalled.

A secondary instrumental cause was the stall indicator system itself. Counter-intuitively, the stall warning was issued when the aircraft was nosed down, and stopped when the aircraft nosed up. This was due to the flight computer treating the extreme AOA calculations (in excess of 40 degrees) as unreliable. Nosing down brought the AOA back into the "reliable" band, but the resulting alarms were counter-intuitive to the pilots; applying a control input and hearing an audible warning is usually a signal to back off that control input.
Finally, the flight crew themselves demonstrated a lack of familiarity with the behavior of the aircraft in exceptional circumstances; the loss of pitch/roll limiters and the counter-intuitive stall horn behavior caused the flight crew, in the absence of the captain, to make a series of critical and ultimately fatal overcorrections and inactions.

Don't forget the middle part where the pilot flying botched the unreliable airspeed procedure. The procedure calls for maintaining certain fixed pitch and maintaining altitude with thrust, but the pilot let it climb which is how it got stalled.

Thanks for that, Keith S. I agree totally with what you say. My personal opinion (for what it's worth after 60+ years in aviation) is that aft CG limits control in some circumstances. Tip stall with swept wings aggravates the situation.

Finally, the flight crew themselves demonstrated a lack of familiarity with the behavior of the aircraft in exceptional circumstances Although all pilots are trained to handle stalls I would contend that few have experienced an actual stall in an airliner. Fewer yet have experienced a high altitude, high mach stall in imc. The pf apparently mistook the stall buffeting for overspeed buffeting as he referred a couple of times to a "crazy speed." But obviously as @JanHudec said, they didn't follow the procedures or they would never have stalled in the first place.

I was mainly talking about the backup flight computer modes. With most of the limiters disabled, the aircraft accepted input it would have otherwise rejected, and became much more sensitive to said input which kept the pilot occupied keeping her level while not paying attention to other elements of her flight envelope.

@KeithS The pilots weren't keeping it level, though. And all of the flight deck instruments were fully operational and indicating correctly for the last 4 minutes of flight; the pitot tubes had cleared of ice. Furthermore, airliners have multiple sources of speed information. While GPS-indicated speed isn't equal to airspeed, when it says you're moving at 60 knots, you can generally assume that the wind around you isn't Mach 0.9 and, thus, you aren't supersonic. More information is available in a translation of the CVR annotated with FDR data.


text:
What is this aircraft from "Devil Girl From Mars" (1954)?
This view is cropped from the opening seconds of the film Devil Girl From Mars. Within the film narrative, this one was supposedly carrying passengers over the Scottish Highlands in the then-present-day, which turns out to be 1954.

If you can help me with the type, I can find most of the other information I need by myself.  However, I would also be interested to know

(a) whether there is anything to be discovered from the barely-visible markings, and
(b) if it's possible to tell whether this shot represents a particular model of the type in question, and what that might say about the background of this specific machine (regardless of context in the film).

It's a Vickers VC.1 Viking in the livery of British European Airways (BEA).

(By RuthAS (Own work) [CC BY 3.0 (http://creativecommons.org/licenses/by/3.0)], via Wikimedia Commons)

The particular type would be 610, the type built for BEA, according to Wikipedia.


text:
Why does the F-22's wing give it such great maneuverability?
An F-22's wing is much different than a wing like the one on the F-16.
This an F-16 wing design:

(Source: www.aviastar.org).
This is an F-22 wing:

(Source: s3.zetaboards.com)
There is a great difference between the 2 wing designs. The F-22 is a very maneuverable aircraft and can out perform almost all the current fighter jets. The YF-23 has a very similar wing design and is considered to be one of the most maneuverable aircraft of all time.
So why does this wing design give the F-22 and YF-23 such good maneuverability?
The YF-23:

(Source: en.wikipedia.org)

You are assuming that the wing design is the reason for the F-22A's maneuverability. A better question would simply ask what makes the aircraft as maneuverable as it is.

I think (I'm happy to be wrong) the consensus is actually that the YF-23 was not as maneuverable, due to the V-tail.

The wing design is different between F-16 and F-22 (which is similar to YF-23) because the requirements were different.
The wing of F-22 (and that of F-23) were designed with stealth in mind, with the wing (and control surface) edges parallel to each other. There was no such design requirement in case of F16.
F-16 is a highly maneuverable aircraft in its own right. However, there are some reasons the F-22 is more maneuverable (i.e. it is super maneuverable) compared to the F-16, most of which are unrelated to wing design.

In general, lesser the wing loading, higher the performance. F22 has a lesser wing loading compared to F-16 (377 kg/m² vs 431 kg/m²). Though this figure will vary depending on various conditions, the F-22 has a lower value, helping in maneuverability.
The F22 has thrust vectoring, which basically acts to increase the control authority and also helps in controlling in high alpha region where the conventional controls would've become unusable.
The F22 is maneuverable in the post-stall region, something the F-16 is not capable of. This helps the F-22 in performing maneuvers like the Cobra and Kulbit, which is not possible in F-16.

As a side note, the F-16 VISTA technology demonstrator (which had thrust vectoring) is considered super-maneuverable on par with THE F-22, though it had the same wing.

I like this answer. It should be stated that what's best for stealth and what's best for maneuverability are generally exclusive of each other (consider the F-117, nicknamed the "Wobbly Goblin"), and the F-22 is maneuverable in spite of its stealth more than anything else. The YF-23 was actually more stealthy than the Raptor, but it paid for it with less agility (pitch rate was reduced due to the rudder-vaters not acting in the pure pitch plane), and as the ability to fight was considered more important than the ability to hide for the ATF mission profile, the Raptor got the nod.

Is point 3 due to anything else besides thrust vectoring?

@zymhan: No. The rest of this line is only here to fulfil the minimum length requirement.

Food for thought-- what were the benefits of the F16XL with its much larger wing, compared to the standard wing?  https://en.wikipedia.org/wiki/General_Dynamics_F-16XL

@quietflyer would make a nice question in its own right. Main reason at the time was larger fuel capacity for longer range and more place for carrying stores.